Question

The phase difference between applied emf and the line current in an anti resonant circuit at resonance is
A. \[\dfrac{\pi }{2}{\rm{ radian}}\]
B. \[\pi {\rm{ radian}}\]
C. \[\dfrac{{3\pi }}{2}{\rm{ radian}}\]
D. zero

Answer 1

Hint: In the solution we will be using the expression of phase difference given in the form of impedance of the circuit. Impedance is the term which takes into account both inductive reactance and capacitive reactance of the given circuit. We will be substituting the value of the difference between reactance of capacitor and inductor to zero at resonance of the given L-R-C circuit.

Complete step by step solution:
The expression for phase difference in an L-R-C circuit.
\[\tan \phi = \dfrac{{\left( {{X_L} - {X_C}} \right)}}{R}\]……(1)

Here, R is the resistance, \[{X_L}\] and \[{X_C}\] are inductive and capacitive reactance respectively.

On rearranging equation (1) we get,
\[\phi = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {{X_L} - {X_C}} \right)}}{R}} \right]\]……(2)

In the phenomena of resonance, the magnitude of reactance of both inductor and capacitor becomes equal but their direction is opposite that means both the reactance will cancel out each other. In other words it can be said that at resonance of an L-R-C circuit the value of the difference between reactance of inductor and capacitor is zero that is \[\left( {{X_L} - {X_C}} \right)\] equal to zero.

Substitute \[0\] for \[{X_L} - {X_C}\] in equation (2).
\[\begin{array}{c}
\phi = {\tan ^{ - 1}}\left[ {\dfrac{0}{R}} \right]\\
 = 0
\end{array}\]

Therefore, the phase difference between applied emf and the line current is zero and the option (D) is correct.

Note: Keep in mind that the value of phase difference \[\phi \] has to be evaluated so after calculating the value of \[\tan \phi \] take the inverse of this value so that the value of phase difference will come out.